Solving Initial Value ODE: x^2y''+xy'+y=0

[freezer]

Homework Statement

Solve the initial value problem:

[tex]
x^{2}{y}'' + x{y}' + y = 0, x>0, y(1)=1, {y}'=2
[/tex]

Homework Equations

y=x^m

The Attempt at a Solution

[tex]
x^{2}(m(m-1)x^{m-2})+xmx^{m-1} + x^{m}
[/tex]

[tex]
x^{2}(m(m-1)x^{m-2})+xmx^{m-1} + x^{m}
[/tex]

[tex]
x^{m}(m(m-1) + m + 1)
[/tex]

[tex]
m = \pm i
[/tex]

This is the way i was doing it:
[tex]
C_1 e^{it} + C_2 e^{-it}
[/tex]

[tex]
C_1(cos(t) + i sin(t)) + C_2(cos(t) - i sin(t))
[/tex]


The solution shows:
[tex]
C_1 x^{i} + C_2 x^{-i}
[/tex]
[tex]
C_1(cos(ln(x)) + i sin(ln(x))) + C_2(cos(ln(x)) - i sin(ln(x)))
[/tex]


With the initial conditions indicate that the solution is correct. Yet the textbook shows the form to be:

[tex]

Y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}

[/tex]

not

[tex]

Y(x) = C_1 x^{r_1} + C_2 x^{r_2}
[/tex]

And if the second form is correct, I have not found an Euler identity that supports the step.

 

[Mark44]

Homework Statement

Solve the initial value problem:

[tex]
x^{2}{y}'' + x{y}' + y = 0, x>0, y(1)=1, {y}'=2
[/tex]

Homework Equations

y=x^m

The Attempt at a Solution

[tex]
x^{2}(m(m-1)x^{m-2})+xmx^{m-1} + x^{m}
[/tex]

[tex]
x^{2}(m(m-1)x^{m-2})+xmx^{m-1} + x^{m}
[/tex]

[tex]
x^{m}(m(m-1) + m + 1)
[/tex]

[tex]
m = \pm i
[/tex]

This is the way i was doing it:
[tex]
C_1 e^{it} + C_2 e^{-it}
[/tex]

Your assumption was that y = x^m was a solution. Using that assumption, you solved for m and got ±i.

[tex]
C_1(cos(t) + i sin(t)) + C_2(cos(t) - i sin(t))
[/tex]The solution shows:
[tex]
C_1 x^{i} + C_2 x^{-i}
[/tex]
[tex]
C_1(cos(ln(x)) + i sin(ln(x))) + C_2(cos(ln(x)) - i sin(ln(x)))
[/tex]With the initial conditions indicate that the solution is correct. Yet the textbook shows the form to be:

[tex]

Y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}

[/tex]

not

[tex]

Y(x) = C_1 x^{r_1} + C_2 x^{r_2}
[/tex]

And if the second form is correct, I have not found an Euler identity that supports the step.

I'm confused. You say that

The solution shows:
[tex]C_1 x^{i} + C_2 x^{-i}[/tex]
[tex]C_1(cos(ln(x)) + i sin(ln(x))) + C_2(cos(ln(x)) - i sin(ln(x)))[/tex]

and then you say

Yet the textbook shows the form to be:

[tex]Y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}[/tex]

Which of these does your book show as the solution?

If the solution you found (i.e., y = c₁xⁱ + c₂x^-i) satisfies the differential equation, then it is the general solution. From the initial conditions you can find the constants c₁ and c₂.

 

[vanhees71]

You have the correct solution. From your findings of two linearly independent solutions
[tex]y_1(x)=\exp(\mathrm{i} \ln x), \quad y_2(x)=\exp(-\mathrm{i} \ln x).[/tex]
The general solution is given by
[tex]y(x)=C_1 y_1(x) + C_2 y_2(x).[/tex]
You can also choose any other two linearly independent solutions, e.g.,
[tex]\tilde{y}_1(x)=\frac{1}{2} [y_1(x)+y_2(x)]=\cos(\ln x), \quad \tilde{y}_1(x) = \frac{1}{2 \mathrm{i}} [y_1(x)+y_2(x)]=\sin(\ln x).[/tex]
The general solution is again given by all linear combinations of these two functions.

 

[freezer]

Okay, where does the ln x come from?

 

[freezer]

Your assumption was that y = x^m was a solution. Using that assumption, you solved for m and got ±i.


I'm confused. You say that


and then you say


Which of these does your book show as the solution?

If the solution you found (i.e., y = c₁xⁱ + c₂x^-i) satisfies the differential equation, then it is the general solution. From the initial conditions you can find the constants c₁ and c₂.

The textbook shows in the examples the e^{rt} while the solution to this problem shows x^r and there is nothing in the text of the chapter that shows this process. I was able to take cos(ln(x) + i sin(ln(x) back to x^i in wolfram but not sure what i am missing where the ln(x) is coming from and the identity for the x^i.

 

[Mark44]

The textbook shows in the examples the e^{rt} while the solution to this problem shows x^r and there is nothing in the text of the chapter that shows this process. I was able to take cos(ln(x) + i sin(ln(x) back to x^i in wolfram but not sure what i am missing where the ln(x) is coming from and the identity for the x^i.

x^m = (e^ln(x))^m = e^m*ln(x). This holds as long as x > 0.

 

[freezer]

x^m = (e^ln(x))^m = e^m*ln(x). This holds as long as x > 0.

Sure e^{ln(x)} = x, but where did the ln(x) come from. I have went through the chapters in the book and I have not found a single theorem or example that shows this.

The examples from the chapter show that when you get your zeros, you use e^{rt). In this case in place of t they used ln(x). I just don't see where the ln(x) came from or what basis this was done with this problem.

Even over at Pauls notes does not show this:
http://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx

 

[LCKurtz]

Perhaps your text or teacher has talked about how the equation ##ax^2y''+bxy' + cy = 0## can be converted to a similar constant coefficent equation ##a\ddot y + b\dot y + cy = 0## by the change of variable ##t=\ln x##. Then you solve the second equation for y as a function of ##t## with your usual ##e^{rt}## method. Once you have that solution you put ##t=\ln x## in and presto, you have the solution to the original problem. The method you describe sounds like that is what they are doing.

 

[ehild]

Homework Statement

Solve the initial value problem:

[tex]
x^{2}{y}'' + x{y}' + y = 0, x>0, y(1)=1, {y}'=2
[/tex]

Homework Equations

y=x^m

The Attempt at a Solution

...
[tex]
m = \pm i
[/tex]

This is the way i was doing it:
[tex]
C_1 e^{it} + C_2 e^{-it}
[/tex]

You did not specify what t is. y is function of x. You assumed the solutions in form [tex]y=x^m [/tex] and you got

[tex]
m = \pm i
[/tex]

The solutions are y₁=xⁱ and y₂=x^-i, the general solution is y=C₁xⁱ+C₂x^-i.

[tex]x=e^{\ln(x)}[/tex].

With that, you can write the solution also in the form
[tex]
y=C_1 e^{i\ln(x)}+ C_2 e^{-i\ln(x)}
[/tex]ehild

 

[LCKurtz]

Perhaps your text or teacher has talked about how the equation ##ax^2y''+bxy' + cy = 0## can be converted to a similar constant coefficent equation ##a\ddot y + b\dot y + cy = 0## by the change of variable ##t=\ln x##. Then you solve the second equation for y as a function of ##t## with your usual ##e^{rt}## method. Once you have that solution you put ##t=\ln x## in and presto, you have the solution to the original problem. The method you describe sounds like that is what they are doing.

I noticed a typo after it was too late to edit. The equation ##a\ddot y + b\dot y + cy = 0## should have been ##A\ddot y + B\dot y + Cy = 0##. The transformed equation has different constants than the original.